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JEE Mains · Maths · STD 12 - 11. three dimension geometry

माना समतल \(\mathrm{P}: 8 \mathrm{x}+\alpha_1 \mathrm{y}+\alpha_2 \mathrm{z}+12=0\), रेखा \(\mathrm{L}: \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}\) के समांतर है। यदि \(\mathrm{P}\) का \(\mathrm{y}\)-अक्ष पर अंतःखंड \(1\) है, तो \(\mathrm{P}\) तथा \(\mathrm{L}\) के बीच दूरी है :

  1. A \(\sqrt{14}\)
  2. B \(\frac{6}{\sqrt{14}}\)
  3. C \(\sqrt{\frac{2}{7}}\)
  4. D \(\sqrt{\frac{7}{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{14}\)

Step-by-step Solution

Detailed explanation

P: \(8 x+\alpha_1 y+\alpha_2 z+12=0\) \(L : \frac{ x +2}{2}=\frac{ y -3}{3}=\frac{ z +4}{5}\) \(\because\) \(P\) is parallel to \(L\) \(\Rightarrow 8(2)+\alpha_1(3)+5\left(\alpha_2\right)=0\) \(\Rightarrow 3 \alpha_1+5\left(\alpha_2\right)=-16\) Also \(y\)-intercept of plane…
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